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devel / comp.theory / Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

SubjectAuthor
* Black box halt decider is NOT a partial deciderMr Flibble
`* Black box halt decider is NOT a partial deciderChris M. Thomasson
 `* Black box halt decider is NOT a partial deciderDavid Brown
  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   +* Black box halt decider is NOT a partial deciderRichard Damon
   |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   | `* Black box halt decider is NOT a partial deciderRichard Damon
   |  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   +- Black box halt decider is NOT a partial deciderRichard Damon
   |   +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | | +- Black box halt decider is NOT a partial deciderRichard Damon
   |   | | `* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   +* Black box halt decider is NOT a partial deciderAndré G. Isaak
   |   | |   |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   | `* Black box halt decider is NOT a partial deciderMike Terry
   |   | |   |  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |   `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    +- Black box halt decider is NOT a partial deciderMike Terry
   |   | |   |    +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |   |    |+* Black box halt decider is NOT a partial deciderJeff Barnett
   |   | |   |    ||+- Black box halt decider is NOT a partial deciderJeff Barnett
   |   | |   |    ||`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |   |    || +- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    || `* Black box halt decider is NOT a partial deciderJeff Barnett
   |   | |   |    ||  `- Black box halt decider is NOT a partial deciderMike Terry
   |   | |   |    |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    | `* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |   |    |  `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    |   +- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    |   `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    |    `- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   |    `- Black box halt decider is NOT a partial deciderwij
   |   | |   +* Black box halt decider is NOT a partial deciderRichard Damon
   |   | |   |`* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   | `* Black box halt decider is NOT a partial deciderRichard Damon
   |   | |   |  `- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |   `* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |    +* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |    |`* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |    | `* Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |    |  `* Black box halt decider is NOT a partial deciderRichard Damon
   |   | |    |   `- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | |    `* Black box halt decider is NOT a partial deciderAndré G. Isaak
   |   | |     +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |     |+- Black box halt decider is NOT a partial deciderAndré G. Isaak
   |   | |     |`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |     | +* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |     | |+* Black box halt decider is NOT a partial deciderAndy Walker
   |   | |     | ||`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |     | || +* Black box halt decider is NOT a partial deciderMalcolm McLean
   |   | |     | || |+* Black box halt decider is NOT a partial decider [ H(P,P)==0 is always correct ]olcott
   |   | |     | || ||`- Black box halt decider is NOT a partial decider [ H(P,P)==0 isRichard Damon
   |   | |     | || |+* Black box halt decider is NOT a partial decider [ H(P,P)==0 is always correct ]olcott
   |   | |     | || ||+- Black box halt decider is NOT a partial decider [ H(P,P)==0 isAndré G. Isaak
   |   | |     | || ||+* Black box halt decider is NOT a partial decider [ H(P,P)==0 isRichard Damon
   |   | |     | || |||`* Black box halt decider is NOT a partial decider [ H(P,P)==0 isMalcolm McLean
   |   | |     | || ||| `* Black box halt decider is NOT a partial decider [ H(P,P)==0 isRichard Damon
   |   | |     | || |||  `- Black box halt decider is NOT a partial decider [ H(P,P)==0 isJeff Barnett
   |   | |     | || ||`- Black box halt decider is NOT a partial decider [ H(P,P)==0 is always correct ]Ben Bacarisse
   |   | |     | || |+* Black box halt decider is NOT a partial deciderBen Bacarisse
   |   | |     | || ||`* Black box halt decider is NOT a partial deciderMalcolm McLean
   |   | |     | || || `* Black box halt decider is NOT a partial decider [ paradox ratherolcott
   |   | |     | || ||  +- Black box halt decider is NOT a partial decider [ paradox ratherRichard Damon
   |   | |     | || ||  `* Black box halt decider is NOT a partial decider [ paradox ratherAndré G. Isaak
   |   | |     | || ||   `* Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]olcott
   |   | |     | || ||    +- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || ||    `* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||     `* Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]olcott
   |   | |     | || ||      +* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||      |`* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||      | `- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || ||      `* Black box halt decider is NOT a partial decider [ H refutesJeff Barnett
   |   | |     | || ||       `* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||        `* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||         +* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||         |+- Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||         |`- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || ||         `* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||          +* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||          |`* Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]olcott
   |   | |     | || ||          | `* Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||          |  `* Black box halt decider is NOT a partial decider [ H refutesolcott
   |   | |     | || ||          |   +- Black box halt decider is NOT a partial decider [ H refutesAndré G. Isaak
   |   | |     | || ||          |   +- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || ||          |   `* _Black_box_halt_decider_is_NOT_a_partial_decider_[_André_doesn't_know_Rice's_Theolcott
   |   | |     | || ||          |    +* _Black_box_halt_decider_is_NOT_a_partial_decider_[André G. Isaak
   |   | |     | || ||          |    |`* _Black_box_halt_decider_is_NOT_a_partial_decider_[olcott
   |   | |     | || ||          |    | +* _Black_box_halt_decider_is_NOT_a_partial_decider_[André G. Isaak
   |   | |     | || ||          |    | |`* _Black_box_halt_decider_is_NOT_a_partial_decider_Malcolm McLean
   |   | |     | || ||          |    | | `* _André_doesn't_know_Rice's_Theorem_[_Malcolm_]olcott
   |   | |     | || ||          |    | |  +* _André_doesn't_know_Rice's_Theorem_[_MalcRichard Damon
   |   | |     | || ||          |    | |  |`* _André_doesn't_know_Rice's_Theorem_[_Malcolcott
   |   | |     | || ||          |    | |  | `* _André_doesn't_know_Rice's_Theorem_[_MalcRichard Damon
   |   | |     | || ||          |    | |  |  `* _André_doesn't_know_Rice's_Theorem_[_Malcolm_](_attention_deficit_disorder_)olcott
   |   | |     | || ||          |    | |  |   `* _André_doesn't_know_Rice's_Theorem_[_MalcRichard Damon
   |   | |     | || ||          |    | |  |    `* _André_doesn't_know_Rice's_Theorem_[_Malcolcott
   |   | |     | || ||          |    | |  |     +- _André_doesn't_know_Rice's_Theorem_[_MalcRichard Damon
   |   | |     | || ||          |    | |  |     +* _André_doesn't_know_Rice's_Theorem_[_Malcolm_](_attention_deficit_disorder_)olcott
   |   | |     | || ||          |    | |  |     `* André doesn't know Rice's Theorem [ MalcolmBen Bacarisse
   |   | |     | || ||          |    | |  +* _André_doesn't_know_Rice's_Theorem_[_MalcAndré G. Isaak
   |   | |     | || ||          |    | |  `- _André_doesn't_know_Rice's_Theorem_[_MalcJeff Barnett
   |   | |     | || ||          |    | +- _Black_box_halt_decider_is_NOT_a_partial_decider_[Richard Damon
   |   | |     | || ||          |    | `* _Black_box_halt_decider_is_NOT_a_partial_decider_[_André_doesn't_know_Rice's_Theolcott
   |   | |     | || ||          |    `- _Black_box_halt_decider_is_NOT_a_partial_decider_[Richard Damon
   |   | |     | || ||          `- Black box halt decider is NOT a partial decider [ H refutesRichard Damon
   |   | |     | || |`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |     | || `- Black box halt decider is NOT a partial deciderAndy Walker
   |   | |     | |`* Black box halt decider is NOT a partial deciderMike Terry
   |   | |     | `* Black box halt decider is NOT a partial deciderwij
   |   | |     `- Black box halt decider is NOT a partial deciderChris M. Thomasson
   |   | `* Black box halt decider is NOT a partial deciderRichard Damon
   |   `* Black box halt decider is NOT a partial deciderMalcolm McLean
   `* Black box halt decider is NOT a partial deciderJeff Barnett

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Re: Black box halt decider is NOT a partial decider

<sdmruo$bam$1@dont-email.me>

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https://www.novabbs.com/devel/article-flat.php?id=18968&group=comp.theory#18968

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: jbb...@notatt.com (Jeff Barnett)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
Date: Mon, 26 Jul 2021 11:38:25 -0600
Organization: A noiseless patient Spider
Lines: 79
Message-ID: <sdmruo$bam$1@dont-email.me>
References: <20210719214640.00000dfc@reddwarf.jmc>
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 by: Jeff Barnett - Mon, 26 Jul 2021 17:38 UTC

On 7/26/2021 11:08 AM, Andy Walker wrote:
> On 26/07/2021 15:36, Ben Bacarisse wrote:
>> In 1979 CS degrees were not that common, but I also suspect there was a
>> measure of snobbery involved -- CS was an upstart subject and could not,
>> surely, fill a whole honours degree.
>
>     There's a large element of truth in that.  The last bit, that is.
> To put it differently, computers then were mostly large, expensive things
> that occupied large rooms if not entire buildings.  Yeah, there were mini
> computers, but they were widely regarded as toys.  Playing games or music
> or videos on computers was largely forbidden.  The whole point of a
> computer was to solve problems -- differential equations, things with
> large matrices, simulations, timetabling, big data, that sort of thing.
> So you needed (a) engineers who could install computers and keep them
> running, (b) programmers who could understand a real-world problem and
> turn it into computerese, and (c) people who could write compilers and
> operating systems.  For (a) you needed a substantial dollop of advanced
> engineering.  For (b) you needed a dollop of university maths/physics/
> whatever to enable you to understand the problems and how to solve them
> in computer terms.  For (c) you needed substantial experience of
> advanced computing.  None of that was really "a whole honours degree",
> esp given [in those days] a virtual absence of textbooks*.
>
>     It made more sense to set up a joint degree, eg maths and CS.
> We [maths, Nott'm] had enough computing modules for that.  But in the
> end we were overtaken by politics.  We "had" to have a full CS dept
> and course, like it or not, or be left behind by our rivals.  Once
> you have such a dept/course, then the content expands to fill the
> available time;  but that's another matter.  Computers today bear
> little relationship to those of the 1970s, making it more sensible
> to have full CS courses;  and graduates thereof who know virtually
> nothing about maths or physics or anything else beyond school level.
>
> ____
>  * There was a chicken-egg problem.  In the absence, or only recent
>    establishment, of proper CS courses, few undergraduate CS texts
>    were produced or sold -- no market for them.  In the absence of
>    proper texts, it was harder to set up good CS courses.

The USA in the same time frame had sprouted CS programs at most
universities, both public and private. Further, there had developed
strong ties nationally and an interesting community emerged during the
1970s. That community, from the 1960s on, included Europeans and soon
Asians too; there were also research islands in Mexico mostly manned by
US expats. The delay in Asia interactions was mostly political: their
researchers could not talk or travel to the West. BTW, this community
was not just an early US invention, there was strong early forces from
your cold countries too.

I'm somewhat surprised to hear in this thread that you all believe that
your real entry into the CS world occurred so late. My impression from
this side of the pond was that the British Isles were well on there way
by then. Perhaps we had better funding in place and that certainly helps
but money can't necessarily buy interests from smart guys.
--
Jeff Barnett

Re: Black box halt decider is NOT a partial decider

<878s1tapvq.fsf@bsb.me.uk>

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https://www.novabbs.com/devel/article-flat.php?id=18970&group=comp.theory#18970

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
Date: Mon, 26 Jul 2021 19:31:21 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Mon, 26 Jul 2021 18:31 UTC

Jeff Barnett <jbb@notatt.com> writes:

> The USA in the same time frame had sprouted CS programs at most
> universities, both public and private. Further, there had developed
> strong ties nationally and an interesting community emerged during the
> 1970s. That community, from the 1960s on, included Europeans and soon
> Asians too; there were also research islands in Mexico mostly manned
> by US expats. The delay in Asia interactions was mostly political:
> their researchers could not talk or travel to the West. BTW, this
> community was not just an early US invention, there was strong early
> forces from your cold countries too.
>
> I'm somewhat surprised to hear in this thread that you all believe
> that your real entry into the CS world occurred so late. My impression
> from this side of the pond was that the British Isles were well on
> there way by then. Perhaps we had better funding in place and that
> certainly helps but money can't necessarily buy interests from smart
> guys.

I'm not sure what you've taken from the thread that suggests some sort
of late entry. Your impression is that the UK was well on the way, but
I'm not sure where to, so I can't confirm or deny. I don't think the UK
was noticeably behind the curve in CS education.

--
Ben.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

<ZNCdndMEKogNmGL9nZ2dnUU7-KvNnZ2d@giganews.com>

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc>
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From: NoO...@NoWhere.com (olcott)
Date: Mon, 26 Jul 2021 13:57:20 -0500
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 by: olcott - Mon, 26 Jul 2021 18:57 UTC

On 7/26/2021 12:16 PM, Jeff Barnett wrote:
> On 7/26/2021 10:41 AM, olcott wrote:
>> On 7/26/2021 11:09 AM, André G. Isaak wrote:
>
> <SNIP>
>
>>> You're claiming that an answer which does *not* correspond to the
>>> actual answer to the question is somehow 'more correct' than the one
>>> which does correspond to the actual answer to the question.
>>>
>>> That's what I would call a 'pathological claim'.
>>>
>>> André
>>>
>>
>> P(P) != H(P,P) recognizes the pathological self-reference(Olcott 2004)
>> error thus refuting Rice's theorem.
>
> Reread the above and note how polite and nurturing Andre has been. And
> yet you reject his simple, near trivial point. Of course he is correct.
> You should apologize to him, not make silly arguments.
>
> oh yes, how was Andres so polite to you? Well he called it a
> "pathological claim"; a "pathological and irrational claimer" is much
> nearer the mark others would say.
>
> And what do you know about Rice? Since you likely can't boil water
> without supervision to keep you from harm, you couldn't prepare rice. I
> know it's a silly play on words but it's hard to believe that you can
> function in the real world with such a disoriented mind.

The HP has the same self-contradictory pattern as the Liar Paradox.
"This sentence is not true." is indeed not true just
like the P of int main() { P(P); } definitely reaches its halting state.

It is a mistake to take the fact that "This sentence is not true" is not
true as an indication that it is true.

In the exact same way it is a mistake to take the fact that the P of int
main() { P(P); } definitely reaches its halting state as an indication
that it is a halting computation.

_P()
[00000c36](01) 55 push ebp
[00000c37](02) 8bec mov ebp,esp
[00000c39](03) 8b4508 mov eax,[ebp+08] // 2nd Param
[00000c3c](01) 50 push eax
[00000c3d](03) 8b4d08 mov ecx,[ebp+08] // 1st Param
[00000c40](01) 51 push ecx
[00000c41](05) e820fdffff call 00000966 // call H(P,P)
[00000c46](03) 83c408 add esp,+08
[00000c49](02) 85c0 test eax,eax
[00000c4b](02) 7402 jz 00000c4f
[00000c4d](02) ebfe jmp 00000c4d
[00000c4f](01) 5d pop ebp
[00000c50](01) c3 ret
Size in bytes:(0027) [00000c50]

machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
Begin Local Halt Decider Simulation at Machine Address:c36
[00000c36][002117ca][002117ce] 55 push ebp
[00000c37][002117ca][002117ce] 8bec mov ebp,esp
[00000c39][002117ca][002117ce] 8b4508 mov eax,[ebp+08]
[00000c3c][002117c6][00000c36] 50 push eax // push P
[00000c3d][002117c6][00000c36] 8b4d08 mov ecx,[ebp+08]
[00000c40][002117c2][00000c36] 51 push ecx // push P
[00000c41][002117be][00000c46] e820fdffff call 00000966 // call H(P,P)

[00000c36][0025c1f2][0025c1f6] 55 push ebp
[00000c37][0025c1f2][0025c1f6] 8bec mov ebp,esp
[00000c39][0025c1f2][0025c1f6] 8b4508 mov eax,[ebp+08]
[00000c3c][0025c1ee][00000c36] 50 push eax // push P
[00000c3d][0025c1ee][00000c36] 8b4d08 mov ecx,[ebp+08]
[00000c40][0025c1ea][00000c36] 51 push ecx // push P
[00000c41][0025c1e6][00000c46] e820fdffff call 00000966 // call H(P,P)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

It is very easy to see that the pure simulation of the input to H(P,P)
cannot possibly ever reach its final state by the fact that there are no
conditional instructions that can possibly break out of the above
infinite recursion.

It looks like people greatly prefer to be God damned liars on this issue.

I would hope for their sake that the following verse is not literally
true. If it was up to me they would simply be forgiven rather than
condemned to Hell (the literal meaning of: "God damned").

Revelation 21:8 (KJV)
....all liars, shall have their part in the lake which burneth with fire
and brimstone: which is the second death.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
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From: NoO...@NoWhere.com (olcott)
Date: Mon, 26 Jul 2021 14:40:06 -0500
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 by: olcott - Mon, 26 Jul 2021 19:40 UTC

On 7/26/2021 12:00 PM, André G. Isaak wrote:
> On 2021-07-26 10:41, olcott wrote:
>> On 7/26/2021 11:09 AM, André G. Isaak wrote:
>
>>> But H(P, P) *isn't* verified as correct.
>>
>> While the input to H(P,P) is simulated in pure simulation mode it
>> cannot possibly ever reach a final state thus conclusively proving
>> that this input never halts.
>
> But P(P) is defined as including a copy of H which *isn't* run in 'pure
> simulation mode'.
>

Because it is common knowledge that in any scientific investigation when
we examine the effect of an independent variable[1] on a dependent
variable[2] that any back-channel communication from the dependent
variable[2] to the independent variable[1] corrupts the analysis.

[1] The behavior of P
[2] The halt status evaluation by H

The way to correct for this when an input to the halt decider was
intentionally defined to corrupt this process is to examine the behavior
of the input in pure simulation mode expressly disallowing any
corrupting back-channel communication from H to P.

> You can set the *outermost* H to run in 'pure simulator mode' if you
> want. But you can't change what occurs in P's copy of H (or put it in
> some other 'mode') or you are no longer evaluating P but something else.
>
>>>> Because of the fact that No P ever halts unless H(P,P) aborts the
>>>> simulation of its input H(P,P) is more correct than P(P).
>>>
>>> You really need to reread what you've written above and think
>>> carefully about it.
>>>
>>> You're claiming that an answer which does *not* correspond to the
>>> actual answer to the question is somehow 'more correct' than the one
>>> which does correspond to the actual answer to the question.
>>>
>>> That's what I would call a 'pathological claim'.
>>>
>>> André
>>>
>>
>> P(P) != H(P,P) recognizes the pathological self-reference(Olcott 2004)
>> error thus refuting Rice's theorem.
>
> There is no 'pathological self-reference error'. Nothing in Linz's proof
> involves something which refers at all, let alone something which refers
> to itself.
>

https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation

In the last page of the Linz proof provided in my paper above Linz does
apply Ĥ to its own machine decription. Why lie?

> The only pathological error here is your claim that an incorrect answer
> is 'more correct' than a correct answer.
>
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: Black box halt decider is NOT a partial decider

<sdn36o$vp2$1@dont-email.me>

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From: jbb...@notatt.com (Jeff Barnett)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
Date: Mon, 26 Jul 2021 13:42:09 -0600
Organization: A noiseless patient Spider
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 by: Jeff Barnett - Mon, 26 Jul 2021 19:42 UTC

On 7/26/2021 12:31 PM, Ben Bacarisse wrote:
> Jeff Barnett <jbb@notatt.com> writes:
>
>> The USA in the same time frame had sprouted CS programs at most
>> universities, both public and private. Further, there had developed
>> strong ties nationally and an interesting community emerged during the
>> 1970s. That community, from the 1960s on, included Europeans and soon
>> Asians too; there were also research islands in Mexico mostly manned
>> by US expats. The delay in Asia interactions was mostly political:
>> their researchers could not talk or travel to the West. BTW, this
>> community was not just an early US invention, there was strong early
>> forces from your cold countries too.
>>
>> I'm somewhat surprised to hear in this thread that you all believe
>> that your real entry into the CS world occurred so late. My impression
>> from this side of the pond was that the British Isles were well on
>> there way by then. Perhaps we had better funding in place and that
>> certainly helps but money can't necessarily buy interests from smart
>> guys.
>
> I'm not sure what you've taken from the thread that suggests some sort
> of late entry. Your impression is that the UK was well on the way, but
> I'm not sure where to, so I can't confirm or deny. I don't think the UK
> was noticeably behind the curve in CS education.

Well there were comments about lack of textbooks, emphasis on problem
solving (implied not theory), comment on not "a whole honours degree",
etc. It sounded like a description of a program that wasn't well started
yet. I believed otherwise. Yes computers were expensive and many
institutions hadn't decided whether CS was primarily an extension of EE,
mathematics, both, or neither (not really an academic area). However,
the world as a whole was on its merry way and CS was well established by
the middle-late 1970s. That seemed fairly clear to others and myself on
this side of the pond. Perhaps I misread the inner intent of some of the
comments in this thread that made me think some of you thought otherwise.
--
Jeff Barnett

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

<sdn40f$5ea$1@dont-email.me>

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Date: Mon, 26 Jul 2021 13:55:57 -0600
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Mon, 26 Jul 2021 19:55 UTC

On 2021-07-26 12:57, olcott wrote:

> In the exact same way it is a mistake to take the fact that the P of int
> main() { P(P); } definitely reaches its halting state as an indication
> that it is a halting computation.

You seem to have a serious difficulty processing definitions.

The *definition* of a halting computation is a computation which reaches
one of its final states in a finite number of steps. This refers only to
the *actual* computation.

This definition exhaustively partitions the set of computations into two
mutually-exclusive groups. If we observe that a computation such as P(P)
reaches a final state in a finite number of steps, it goes in the
halting group. Otherwise, it goes into the non-halting group.

The definition is clear and unambiguous.

It does not mention anything about simulators or aborted simulations.

It makes no exceptions for 'pathological' inputs.

It doesn't care *how* they end up in their final state. That is, it
doesn't say "except if the only reason it halted was become some
otherwise infinite chain of simulations was aborted".

The *only* thing it cares about is whether the computation ends up in a
final state after a finite number of steps.

Once we have observed that P(P) *does* reach a final state in a finite
number of steps, we put it in the 'halting' group and we are done.

At this point, there are absolutely no additional considerations which
can have any relevance whatsoever. No arguments, or traces, or claims
about 'pathological' anything will make a difference to the fact that it
has been placed in the halting group and thus *cannot* be a non-halting
computation.

The *only* possible way you can claim that H(P, P)==0 is if you are
using an entirely different definition of 'halting' from the rest of the
world, in which case your argument has no bearing on the halting problem
since that problem uses the standard definition of halting. An
Olcott-Halting Decider is *not* a Halting Decider.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc>
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From: NoO...@NoWhere.com (olcott)
Date: Mon, 26 Jul 2021 15:34:45 -0500
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 by: olcott - Mon, 26 Jul 2021 20:34 UTC

On 7/26/2021 2:55 PM, André G. Isaak wrote:
> On 2021-07-26 12:57, olcott wrote:
>
>> In the exact same way it is a mistake to take the fact that the P of
>> int main() { P(P); } definitely reaches its halting state as an
>> indication that it is a halting computation.
>
> You seem to have a serious difficulty processing definitions.
>
> The *definition* of a halting computation is a computation which reaches
> one of its final states in a finite number of steps. This refers only to
> the *actual* computation.
>
> This definition exhaustively partitions the set of computations into two
> mutually-exclusive groups. If we observe that a computation such as P(P)
> reaches a final state in a finite number of steps, it goes in the
> halting group. Otherwise, it goes into the non-halting group.
>
> The definition is clear and unambiguous.
>
> It does not mention anything about simulators or aborted simulations.
>
> It makes no exceptions for 'pathological' inputs.
>
> It doesn't care *how* they end up in their final state. That is, it
> doesn't say "except if the only reason it halted was become some
> otherwise infinite chain of simulations was aborted".
>
> The *only* thing it cares about is whether the computation ends up in a
> final state after a finite number of steps.
>
> Once we have observed that P(P) *does* reach a final state in a finite
> number of steps, we put it in the 'halting' group and we are done.
>
> At this point, there are absolutely no additional considerations which
> can have any relevance whatsoever. No arguments, or traces, or claims
> about 'pathological' anything will make a difference to the fact that it
> has been placed in the halting group and thus *cannot* be a non-halting
> computation.
>
> The *only* possible way you can claim that H(P, P)==0 is if you are
> using an entirely different definition of 'halting' from the rest of the
> world,

A self-contradictory inputs such as the last page of the Linz proof
where the Linz Ĥ is applied to its own TM description must be handled
differently than inputs that are not self-contradictory.

You can say that when Ĥ is applied to its own TM description that it is
not a case of a TM being applied to its own TM description (an obvious
lie) or you can admit that this is a case of self-reference.

Being able to simply recognize that this input is self-contradictory
refutes Rice's Theorem. If the halt decider returned 2 for invalid input
this refutes Rices Theorem.

// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{ if (H(x, x))
HERE: goto HERE;
}

You can also say that when the input to H does the opposite of whatever
that halt decider decides that it is not doing the opposite of whatever
that halt decider decides. This would be an obvious lie.

Or you can admit that when the input does the opposite of whatever the
halt decider decides that it is contradicting the halt decider.

> in which case your argument has no bearing on the halting problem
> since that problem uses the standard definition of halting. An
> Olcott-Halting Decider is *not* a Halting Decider.
>
> André
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

<sdn7bo$sjr$1@dont-email.me>

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https://www.novabbs.com/devel/article-flat.php?id=18977&group=comp.theory#18977

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Date: Mon, 26 Jul 2021 14:53:10 -0600
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Mon, 26 Jul 2021 20:53 UTC

On 2021-07-26 14:34, olcott wrote:
> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>> On 2021-07-26 12:57, olcott wrote:
>>
>>> In the exact same way it is a mistake to take the fact that the P of
>>> int main() { P(P); } definitely reaches its halting state as an
>>> indication that it is a halting computation.
>>
>> You seem to have a serious difficulty processing definitions.
>>
>> The *definition* of a halting computation is a computation which
>> reaches one of its final states in a finite number of steps. This
>> refers only to the *actual* computation.
>>
>> This definition exhaustively partitions the set of computations into
>> two mutually-exclusive groups. If we observe that a computation such
>> as P(P) reaches a final state in a finite number of steps, it goes in
>> the halting group. Otherwise, it goes into the non-halting group.
>>
>> The definition is clear and unambiguous.
>>
>> It does not mention anything about simulators or aborted simulations.
>>
>> It makes no exceptions for 'pathological' inputs.
>>
>> It doesn't care *how* they end up in their final state. That is, it
>> doesn't say "except if the only reason it halted was become some
>> otherwise infinite chain of simulations was aborted".
>>
>> The *only* thing it cares about is whether the computation ends up in
>> a final state after a finite number of steps.
>>
>> Once we have observed that P(P) *does* reach a final state in a finite
>> number of steps, we put it in the 'halting' group and we are done.
>>
>> At this point, there are absolutely no additional considerations which
>> can have any relevance whatsoever. No arguments, or traces, or claims
>> about 'pathological' anything will make a difference to the fact that
>> it has been placed in the halting group and thus *cannot* be a
>> non-halting computation.
>>
>> The *only* possible way you can claim that H(P, P)==0 is if you are
>> using an entirely different definition of 'halting' from the rest of
>> the world,
>
> A self-contradictory inputs such as the last page of the Linz proof
> where the Linz Ĥ is applied to its own TM description must be handled
> differently than inputs that are not self-contradictory.

Where in the definition of halting do you find anything which supports this?

As I said above, once you've determined that P(P) reaches its final
state *absolutely nothing* else is relevant. It is a halting
computation. Nothing you right below alters this.

> You can say that when Ĥ is applied to its own TM description that it is
> not a case of a TM being applied to its own TM description (an obvious
> lie) or you can admit that this is a case of self-reference.

When you make a copy of something, there is no 'reference' at all. If I
take some piece of code and make a copy of it, neither copy refers to
the other.

If a TM expects a description of a computation on the tape, you don't
place the TM itself on the tape. Just because you give TM X a
description of TM X does not make it 'self-referential'.

And whenever any TM which takes a TM description as an input claims to
be 'universal', that means it must be able handle *every* TM
description, including one that happens to be of itself.

> Being able to simply recognize that this input is self-contradictory
> refutes Rice's Theorem. If the halt decider returned 2 for invalid input
> this refutes Rices Theorem.
>
> // Simplified Linz Ĥ (Linz:1990:319)
> // Strachey(1965) CPL translated to C
> void P(u32 x)
> {
>   if (H(x, x))
>     HERE: goto HERE;
> }
>
> You can also say that when the input to H does the opposite of whatever
> that halt decider decides that it is not doing the opposite of whatever
> that halt decider decides. This would be an obvious lie.
>
> Or you can admit that when the input does the opposite of whatever the
> halt decider decides that it is contradicting the halt decider.

Reread the definition of halting. It makes no mention of 'halt deciders'
whatsoever. Computations are paritioned into halting and non-halting
based solely on the behaviour of the computation itself independent of
any putative halt decider.

Since the halting status of a computation is independent of any
particular halt-decider, claiming that there is something wrong with
this 'input' is meaningless. Its not an input to anything. It is a
computation.

You can pass a *description* of P(P) to some turing machine, in which
case a contradiction arises with P(P) *only* if you pass a description
of it to a *very particular* halt decider (the one from which it is
derived).

But the question 'does P(P) halt' is answerable without mentioning *any*
halt decider.

The question is 'does this computation halt'. Not 'does this computation
halt relative to decider X'.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider

<sdnakg$1nig$1@gioia.aioe.org>

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https://www.novabbs.com/devel/article-flat.php?id=18978&group=comp.theory#18978

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From: peterxpe...@hotmail.com (Peter)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
Date: Mon, 26 Jul 2021 22:49:04 +0100
Organization: Aioe.org NNTP Server
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 by: Peter - Mon, 26 Jul 2021 21:49 UTC

Jeff Barnett wrote:
> On 7/26/2021 12:31 PM, Ben Bacarisse wrote:
>> Jeff Barnett <jbb@notatt.com> writes:
>>
>>> The USA in the same time frame had sprouted CS programs at most
>>> universities, both public and private. Further, there had developed
>>> strong ties nationally and an interesting community emerged during the
>>> 1970s. That community, from the 1960s on, included Europeans and soon
>>> Asians too; there were also research islands in Mexico mostly manned
>>> by US expats. The delay in Asia interactions was mostly political:
>>> their researchers could not talk or travel to the West. BTW, this
>>> community was not just an early US invention, there was strong early
>>> forces from your cold countries too.
>>>
>>> I'm somewhat surprised to hear in this thread that you all believe
>>> that your real entry into the CS world occurred so late. My impression
>>> from this side of the pond was that the British Isles were well on
>>> there way by then. Perhaps we had better funding in place and that
>>> certainly helps but money can't necessarily buy interests from smart
>>> guys.
>>
>> I'm not sure what you've taken from the thread that suggests some sort
>> of late entry.  Your impression is that the UK was well on the way, but
>> I'm not sure where to, so I can't confirm or deny.  I don't think the UK
>> was noticeably behind the curve in CS education.
>
> Well there were comments about lack of textbooks, emphasis on problem
> solving (implied not theory), comment on not "a whole honours degree",
> etc. It sounded like a description of a program that wasn't well started
> yet. I believed otherwise. Yes computers were expensive and many
> institutions hadn't decided whether CS was primarily an extension of EE,
> mathematics, both, or neither (not really an academic area). However,
> the world as a whole was on its merry way and CS was well established by
> the middle-late 1970s. That seemed fairly clear to others and myself on
> this side of the pond. Perhaps I misread the inner intent of some of the
> comments in this thread that made me think some of you thought otherwise.

In my day there was something called a conversion MSc. If one had a
first degree in a numerate subject (1st or 2.2, probably) one could take
an MSc in computing that did not presuppose any first degree exposure to
computing. Do they still exist? Where they (or are they) a good idea?

--
The world will little note, nor long remember what we say here
Abraham Lincoln at Gettysburg

Re: Black box halt decider is NOT a partial decider

<8735s0bu1d.fsf@bsb.me.uk>

 copy mid

https://www.novabbs.com/devel/article-flat.php?id=18979&group=comp.theory#18979

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
Date: Mon, 26 Jul 2021 23:16:14 +0100
Organization: A noiseless patient Spider
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 by: Ben Bacarisse - Mon, 26 Jul 2021 22:16 UTC

Jeff Barnett <jbb@notatt.com> writes:

> On 7/26/2021 12:31 PM, Ben Bacarisse wrote:
>> Jeff Barnett <jbb@notatt.com> writes:
>>
>>> The USA in the same time frame had sprouted CS programs at most
>>> universities, both public and private. Further, there had developed
>>> strong ties nationally and an interesting community emerged during the
>>> 1970s. That community, from the 1960s on, included Europeans and soon
>>> Asians too; there were also research islands in Mexico mostly manned
>>> by US expats. The delay in Asia interactions was mostly political:
>>> their researchers could not talk or travel to the West. BTW, this
>>> community was not just an early US invention, there was strong early
>>> forces from your cold countries too.
>>>
>>> I'm somewhat surprised to hear in this thread that you all believe
>>> that your real entry into the CS world occurred so late. My impression
>>> from this side of the pond was that the British Isles were well on
>>> there way by then. Perhaps we had better funding in place and that
>>> certainly helps but money can't necessarily buy interests from smart
>>> guys.
>>
>> I'm not sure what you've taken from the thread that suggests some sort
>> of late entry. Your impression is that the UK was well on the way, but
>> I'm not sure where to, so I can't confirm or deny. I don't think the UK
>> was noticeably behind the curve in CS education.
>
> Well there were comments about lack of textbooks, emphasis on problem
> solving (implied not theory), comment on not "a whole honours degree",
> etc. It sounded like a description of a program that wasn't well
> started yet.

The first ever CS qualification was the Cambridge Diploma which started
in 1953. I'm pretty sure you would not find any reliable textbooks
then!

In the UK, this set something of a precedent. Lots of universities ran
postgraduate diploma or MSc courses so you could "top-up" a degree in
maths or engineering (or, indeed, pretty much anything) to get a
grounding in CS. Many of these courses were excellent.

Exactly how the development of full undergraduate degree programs panned
out in the UK vs the US I don't know. I suspect the US was a few years
ahead, in terms of numbers, but I might be mistaken.

> I believed otherwise. Yes computers were expensive and
> many institutions hadn't decided whether CS was primarily an extension
> of EE, mathematics, both, or neither (not really an academic
> area). However, the world as a whole was on its merry way and CS was
> well established by the middle-late 1970s. That seemed fairly clear to
> others and myself on this side of the pond.

There was something well established by then, but was it really what
we'd call CS now? I'm not sure.

> Perhaps I misread the
> inner intent of some of the comments in this thread that made me think
> some of you thought otherwise.

I think we're probably just talking about different things.

--
Ben.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Newsgroups: comp.theory
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Mon, 26 Jul 2021 22:29 UTC

On 7/26/21 12:40 PM, olcott wrote:
> On 7/26/2021 12:00 PM, André G. Isaak wrote:
>> On 2021-07-26 10:41, olcott wrote:
>>> On 7/26/2021 11:09 AM, André G. Isaak wrote:
>>
>>>> But H(P, P) *isn't* verified as correct.
>>>
>>> While the input to H(P,P) is simulated in pure simulation mode it
>>> cannot possibly ever reach a final state thus conclusively proving
>>> that this input never halts.
>>
>> But P(P) is defined as including a copy of H which *isn't* run in
>> 'pure simulation mode'.
>>
>
> Because it is common knowledge that in any scientific investigation when
> we examine the effect of an independent variable[1] on a dependent
> variable[2] that any back-channel communication from the dependent
> variable[2] to the independent variable[1] corrupts the analysis.
>
> [1] The behavior of P
> [2] The halt status evaluation by H

But the INDEPENDENT variable is H, and the DEPENDENT variable is H^
(aks P) and its behavior. Note H^ is EXPLICITLY dependent on H, to try
and call it an 'independent variable' is a categorical error.

When studing something, it is very important to get your variables right.

>
> The way to correct for this when an input to the halt decider was
> intentionally defined to corrupt this process is to examine the behavior
> of the input in pure simulation mode expressly disallowing any
> corrupting back-channel communication from H to P.

But P is EXPLICITLY a function of H, treating P as the independent
variable is an error.

>
>> You can set the *outermost* H to run in 'pure simulator mode' if you
>> want. But you can't change what occurs in P's copy of H (or put it in
>> some other 'mode') or you are no longer evaluating P but something else.
>>
>>>>> Because of the fact that No P ever halts unless H(P,P) aborts the
>>>>> simulation of its input H(P,P) is more correct than P(P).
>>>>
>>>> You really need to reread what you've written above and think
>>>> carefully about it.
>>>>
>>>> You're claiming that an answer which does *not* correspond to the
>>>> actual answer to the question is somehow 'more correct' than the one
>>>> which does correspond to the actual answer to the question.
>>>>
>>>> That's what I would call a 'pathological claim'.
>>>>
>>>> André
>>>>
>>>
>>> P(P) != H(P,P) recognizes the pathological self-reference(Olcott
>>> 2004) error thus refuting Rice's theorem.
>>
>> There is no 'pathological self-reference error'. Nothing in Linz's
>> proof involves something which refers at all, let alone something
>> which refers to itself.
>>
>
> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>
>
> In the last page of the Linz proof provided in my paper above Linz does
> apply Ĥ to its own machine decription.  Why lie?

H^ is given a copy of its representation. This is NOT self reference, H
is defined to be able to accept ANY machine input, and thus so is H^, so
H^ is NOT built with self reference. It is only a particular instance
where we give it a COPY of its representation.

This is NOT self-reference like you "Thist statement is false" statement
which EXPLICITLY references itself.

Re: Black box halt decider is NOT a partial decider

<87wnpcadjz.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider
Date: Mon, 26 Jul 2021 23:57:36 +0100
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 by: Ben Bacarisse - Mon, 26 Jul 2021 22:57 UTC

Peter <peterxpercival@hotmail.com> writes:

> In my day there was something called a conversion MSc. If one had a
> first degree in a numerate subject (1st or 2.2, probably) one could
> take an MSc in computing that did not presuppose any first degree
> exposure to computing. Do they still exist? Where they (or are they)
> a good idea?

I think they are much less common than they used to be, but I note with
some pride that the one I used to run is still going. I think it was
the second one started in the UK, after Cambridge's diploma course.
Obviously I think they are (or at least were) a very good idea, but then
I am clearly biased.

From the other side, the students were great -- motivated and
enthusiastic with a ride range of prior experience. We admitted
students fresh out of an undergraduate degree as well as many with
significant careers already under their belts. And we would often take
a punt on good students with unusual degrees and/or backgrounds.

Such courses were usually a full calendar year, with the summer taken up
with a significant (supervised) dissertation. You can fit a lot into a
full 12 month course. (I always hated the "you have the summer off"
myth of academia.)

--
Ben.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Newsgroups: comp.theory
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Mon, 26 Jul 2021 23:01 UTC

On 7/26/21 1:34 PM, olcott wrote:
> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>> On 2021-07-26 12:57, olcott wrote:
>>
>>> In the exact same way it is a mistake to take the fact that the P of
>>> int main() { P(P); } definitely reaches its halting state as an
>>> indication that it is a halting computation.
>>
>> You seem to have a serious difficulty processing definitions.
>>
>> The *definition* of a halting computation is a computation which
>> reaches one of its final states in a finite number of steps. This
>> refers only to the *actual* computation.
>>
>> This definition exhaustively partitions the set of computations into
>> two mutually-exclusive groups. If we observe that a computation such
>> as P(P) reaches a final state in a finite number of steps, it goes in
>> the halting group. Otherwise, it goes into the non-halting group.
>>
>> The definition is clear and unambiguous.
>>
>> It does not mention anything about simulators or aborted simulations.
>>
>> It makes no exceptions for 'pathological' inputs.
>>
>> It doesn't care *how* they end up in their final state. That is, it
>> doesn't say "except if the only reason it halted was become some
>> otherwise infinite chain of simulations was aborted".
>>
>> The *only* thing it cares about is whether the computation ends up in
>> a final state after a finite number of steps.
>>
>> Once we have observed that P(P) *does* reach a final state in a finite
>> number of steps, we put it in the 'halting' group and we are done.
>>
>> At this point, there are absolutely no additional considerations which
>> can have any relevance whatsoever. No arguments, or traces, or claims
>> about 'pathological' anything will make a difference to the fact that
>> it has been placed in the halting group and thus *cannot* be a
>> non-halting computation.
>>
>> The *only* possible way you can claim that H(P, P)==0 is if you are
>> using an entirely different definition of 'halting' from the rest of
>> the world,
>
> A self-contradictory inputs such as the last page of the Linz proof
> where the Linz Ĥ is applied to its own TM description must be handled
> differently than inputs that are not self-contradictory.
>

WHY? H is suppose to be able to handle ANY input.

> You can say that when Ĥ is applied to its own TM description that it is
> not a case of a TM being applied to its own TM description (an obvious
> lie) or you can admit that this is a case of self-reference.
>

H^ is designed to take ANY machine description as a input, Thus it needs
to be able take H^ as a description. So it isn't special to give it a
copy of itself.

> Being able to simply recognize that this input is self-contradictory
> refutes Rice's Theorem. If the halt decider returned 2 for invalid input
> this refutes Rices Theorem.
>

The problem is that H^ HAS a definite Halting value for a given H, and
of H does given an aswer for H(H^,H^) then other halt deciders are able
to give the right answer, so it isn't that you need to 'redefine' what
halting is, but accept that H can't give the right answer for H^.

The problem with your '2' answer is that, yes, if the decider could
actually detect this specific case and return a 'invalid' result, it
might defeat THIS particual proof, but there are many other proofs that
show other machines that H could not get the right answer to. I had
thought about this 'invalid' result, but one BIG issue is this means
that the 'right' answer for the Halting Question is a function of what
decider you ask, as if H^ converts this 'invalid' return to something
definite (either Halting or Non-Halting) then H's right answer to
H(H^,H^) is different than any other decider that would give that
definite answer.

> // Simplified Linz Ĥ (Linz:1990:319)
> // Strachey(1965) CPL translated to C
> void P(u32 x)
> {
>   if (H(x, x))
>     HERE: goto HERE;
> }
>
> You can also say that when the input to H does the opposite of whatever
> that halt decider decides that it is not doing the opposite of whatever
> that halt decider decides. This would be an obvious lie.
>
> Or you can admit that when the input does the opposite of whatever the
> halt decider decides that it is contradicting the halt decider.

Yes, the input does the opposite of what the decider decides, thus, the
decider can never be correct.

What's the problem. If you problem ALLOWS (as the Halting Problme does)
for the thing being decided on to use the decider in generating its
response, then the decider can't be totally accurate. That is basic
logic. If you have to play a game against someone that knows what you
are going to do, it is hard to win.

>
>
>
>> in which case your argument has no bearing on the halting problem
>> since that problem uses the standard definition of halting. An
>> Olcott-Halting Decider is *not* a Halting Decider.
>>
>> André
>>
>
>

Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 is correct ]

<87pmv4ab6r.fsf@bsb.me.uk>

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From: ben.use...@bsb.me.uk (Ben Bacarisse)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 is correct ]
Date: Tue, 27 Jul 2021 00:48:44 +0100
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 by: Ben Bacarisse - Mon, 26 Jul 2021 23:48 UTC

olcott <NoOne@NoWhere.com> writes:

> While the input to H(P,P) is simulated in pure simulation mode it
> cannot possibly ever reach a final state thus conclusively proving
> that this input never halts.

Who cares? H(P,P) == 0 and P(P) halts so H is not a halt decider. You
once claimed something "interesting" i.e. that you had

"encoded all of the ... Linz Turing machine H that correctly decides
halting for its fully encoded input pair: (Ĥ, Ĥ)"

and you insisted that

"Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
specs does not exist. I now have a fully encoded pair of Turing
Machines H / Ĥ proving them wrong."

Does that "Linz Turing machine H" accept or reject the "fully encoded
input pair (H^, H^)", and what is the halting status if H^ when given H^
(encoded)? If, as now, H rejects (H^, H^) but H^ halts when given H^
then there was never anything interesting about what you were claiming,
but it was at least about Turing machines and the proof you have fixated
on.

But you also said your TMs H and H^ are "exactly and precisely as in
Linz", so really either

(1) H rejects (H^, H^) and H^ does not halt on input H^, or
(2) H accepts (H^, H^) and H^ halts on input H^

should be the case. So, come clean. Which was the case back then:

(1) H rejects (H^, H^) and H^ does not halt in input H^, or
(2) H accepts (H^, H^) and H^ halts on input H^, or
(3) H rejects (H^, H^) and H^ halts on input H^.

Without the comfort blanket of your huge pile of junk x86 code, I
suspect you won't dare say.

--
Ben.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
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From: NoO...@NoWhere.com (olcott)
Date: Mon, 26 Jul 2021 19:14:54 -0500
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 by: olcott - Tue, 27 Jul 2021 00:14 UTC

On 7/26/2021 2:55 PM, André G. Isaak wrote:
> On 2021-07-26 12:57, olcott wrote:
>
>> In the exact same way it is a mistake to take the fact that the P of
>> int main() { P(P); } definitely reaches its halting state as an
>> indication that it is a halting computation.
>
> You seem to have a serious difficulty processing definitions.
>
> The *definition* of a halting computation is a computation which reaches
> one of its final states in a finite number of steps. This refers only to
> the *actual* computation.
>
> This definition exhaustively partitions the set of computations into two
> mutually-exclusive groups. If we observe that a computation such as P(P)
> reaches a final state in a finite number of steps, it goes in the
> halting group. Otherwise, it goes into the non-halting group.
>
> The definition is clear and unambiguous.
>
> It does not mention anything about simulators or aborted simulations.
>
> It makes no exceptions for 'pathological' inputs.
>

None-the-less if pathological inputs can be recognized this refutes Rice.
None-the-less if pathological inputs can be recognized this refutes Rice.
None-the-less if pathological inputs can be recognized this refutes Rice.
None-the-less if pathological inputs can be recognized this refutes Rice.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

<sdnjhs$30s$1@dont-email.me>

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Date: Mon, 26 Jul 2021 18:21:15 -0600
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Tue, 27 Jul 2021 00:21 UTC

On 2021-07-26 18:14, olcott wrote:
> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>> On 2021-07-26 12:57, olcott wrote:
>>
>>> In the exact same way it is a mistake to take the fact that the P of
>>> int main() { P(P); } definitely reaches its halting state as an
>>> indication that it is a halting computation.
>>
>> You seem to have a serious difficulty processing definitions.
>>
>> The *definition* of a halting computation is a computation which
>> reaches one of its final states in a finite number of steps. This
>> refers only to the *actual* computation.
>>
>> This definition exhaustively partitions the set of computations into
>> two mutually-exclusive groups. If we observe that a computation such
>> as P(P) reaches a final state in a finite number of steps, it goes in
>> the halting group. Otherwise, it goes into the non-halting group.
>>
>> The definition is clear and unambiguous.
>>
>> It does not mention anything about simulators or aborted simulations.
>>
>> It makes no exceptions for 'pathological' inputs.
>>
>
> None-the-less if pathological inputs can be recognized this refutes Rice.
> None-the-less if pathological inputs can be recognized this refutes Rice.
> None-the-less if pathological inputs can be recognized this refutes Rice.
> None-the-less if pathological inputs can be recognized this refutes Rice.

There *are no* patholological inputs. There are simply inputs that H
gets wrong. There's nothing absolutely 'pathological' about that.

You're entirely missing the point. In what possible sense, given the
definition given above, can P(P) be construed as non-halting? It meets
the definition of halting in every possible way. Yet you insist that
your 'decider' is correct to claim that H(P, P) == 0 is false.

How can this claim be justified *in terms of the definition of halting*
and in terms of the *actual behaviour* of P(P). Those are the only two
things that matter.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc>
<sd8bim$1set$1@gioia.aioe.org> <87eebrlv2m.fsf@bsb.me.uk>
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Tue, 27 Jul 2021 01:03 UTC

On 7/26/21 5:14 PM, olcott wrote:
> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>> On 2021-07-26 12:57, olcott wrote:
>>
>>> In the exact same way it is a mistake to take the fact that the P of
>>> int main() { P(P); } definitely reaches its halting state as an
>>> indication that it is a halting computation.
>>
>> You seem to have a serious difficulty processing definitions.
>>
>> The *definition* of a halting computation is a computation which
>> reaches one of its final states in a finite number of steps. This
>> refers only to the *actual* computation.
>>
>> This definition exhaustively partitions the set of computations into
>> two mutually-exclusive groups. If we observe that a computation such
>> as P(P) reaches a final state in a finite number of steps, it goes in
>> the halting group. Otherwise, it goes into the non-halting group.
>>
>> The definition is clear and unambiguous.
>>
>> It does not mention anything about simulators or aborted simulations.
>>
>> It makes no exceptions for 'pathological' inputs.
>>
>
> None-the-less if pathological inputs can be recognized this refutes Rice.
> None-the-less if pathological inputs can be recognized this refutes Rice.
> None-the-less if pathological inputs can be recognized this refutes Rice.
> None-the-less if pathological inputs can be recognized this refutes Rice.
>

No, because the RULES (i.e. the DEFINITON) of a Halt Decider is that it
has to answer that the machine Halts or that it is Non-Halting.

It doesn't have the OPTION of saying 'invalid' because no input is
actually invalid.

All that detecting the 'pathological' state means is that H KNOWS it
will be wrong with what ever answer it gives.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]
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References: <20210719214640.00000dfc@reddwarf.jmc> <87eebrlv2m.fsf@bsb.me.uk> <sdckqo$cm8$1@gioia.aioe.org> <87a6mehx5q.fsf@bsb.me.uk> <sdfbv2$14bi$2@gioia.aioe.org> <875yx0he2s.fsf@bsb.me.uk> <sdffqm$jsh$1@gioia.aioe.org> <87zgucfux4.fsf@bsb.me.uk> <sdi9vb$r9b$1@dont-email.me> <87eebnfc8c.fsf@bsb.me.uk> <19Kdna-u6-AOSWH9nZ2dnUU78QXNnZ2d@brightview.co.uk> <87tukjdqmi.fsf@bsb.me.uk> <sdjlo4$1oct$1@gioia.aioe.org> <eOadnfv1CNxEEGD9nZ2dnUU78RPNnZ2d@brightview.co.uk> <4826ab33-061b-472e-a1a5-e2ded35ecd82n@googlegroups.com> <87r1fmcgta.fsf@bsb.me.uk> <8978f969-8b53-4535-9bd3-e838818b9755n@googlegroups.com> <dLudnbEJjIhIrGP9nZ2dnUU7-dPNnZ2d@giganews.com> <sdlg2u$tth$1@dont-email.me> <g5Cdndhbg9ImWWP9nZ2dnUU7-S3NnZ2d@giganews.com> <sdmmo7$72r$1@dont-email.me> <_7OdnVI71OcgeGP9nZ2dnUU7-UXNnZ2d@giganews.com> <sdmqm4$2hr$1@dont-email.me> <ZNCdndMEKogNmGL9nZ2dnUU7-KvNnZ2d@giganews.com> <sdn40f$5ea$1@dont-email.me> <IYGdnV9avtZj0mL9nZ2dnUU7-d-dnZ2d@giganews.com> <sdnjhs$30s$1@dont-email.me>
From: NoO...@NoWhere.com (olcott)
Date: Mon, 26 Jul 2021 20:03:31 -0500
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 by: olcott - Tue, 27 Jul 2021 01:03 UTC

On 7/26/2021 7:21 PM, André G. Isaak wrote:
> On 2021-07-26 18:14, olcott wrote:
>> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>>> On 2021-07-26 12:57, olcott wrote:
>>>
>>>> In the exact same way it is a mistake to take the fact that the P of
>>>> int main() { P(P); } definitely reaches its halting state as an
>>>> indication that it is a halting computation.
>>>
>>> You seem to have a serious difficulty processing definitions.
>>>
>>> The *definition* of a halting computation is a computation which
>>> reaches one of its final states in a finite number of steps. This
>>> refers only to the *actual* computation.
>>>
>>> This definition exhaustively partitions the set of computations into
>>> two mutually-exclusive groups. If we observe that a computation such
>>> as P(P) reaches a final state in a finite number of steps, it goes in
>>> the halting group. Otherwise, it goes into the non-halting group.
>>>
>>> The definition is clear and unambiguous.
>>>
>>> It does not mention anything about simulators or aborted simulations.
>>>
>>> It makes no exceptions for 'pathological' inputs.
>>>
>>
>> None-the-less if pathological inputs can be recognized this refutes Rice.
>> None-the-less if pathological inputs can be recognized this refutes Rice.
>> None-the-less if pathological inputs can be recognized this refutes Rice.
>> None-the-less if pathological inputs can be recognized this refutes Rice.
>
> There *are no* patholological inputs. There are simply inputs that H
> gets wrong. There's nothing absolutely 'pathological' about that.

If my system recognizes whatever you want to call it the cases where the
halt decider gets wrong, this refutes Rice.

At this point I suspect that you may think that I am talking about the
breakfast cereal and have no idea about Rice's Theorem.

In computability theory, Rice's theorem states that all non-trivial,
semantic properties of programs are undecidable.
https://en.wikipedia.org/wiki/Rice%27s_theorem

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 is correct ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 is
correct ]
Newsgroups: comp.theory
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From: NoO...@NoWhere.com (olcott)
Date: Mon, 26 Jul 2021 20:06:50 -0500
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 by: olcott - Tue, 27 Jul 2021 01:06 UTC

On 7/26/2021 6:48 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> While the input to H(P,P) is simulated in pure simulation mode it
>> cannot possibly ever reach a final state thus conclusively proving
>> that this input never halts.
>
> Who cares? H(P,P) == 0 and P(P) halts so H is not a halt decider.

When you feed the system self-contradictory input you get an
inconsistent result that can be used to refute Rice.

> You
> once claimed something "interesting" i.e. that you had
>
> "encoded all of the ... Linz Turing machine H that correctly decides
> halting for its fully encoded input pair: (Ĥ, Ĥ)"
>
> and you insisted that
>
> "Everyone has claimed that H on input pair (Ĥ, Ĥ) meeting the Linz
> specs does not exist. I now have a fully encoded pair of Turing
> Machines H / Ĥ proving them wrong."
>
> Does that "Linz Turing machine H" accept or reject the "fully encoded
> input pair (H^, H^)", and what is the halting status if H^ when given H^
> (encoded)? If, as now, H rejects (H^, H^) but H^ halts when given H^
> then there was never anything interesting about what you were claiming,
> but it was at least about Turing machines and the proof you have fixated
> on.
>
> But you also said your TMs H and H^ are "exactly and precisely as in
> Linz", so really either
>
> (1) H rejects (H^, H^) and H^ does not halt on input H^, or
> (2) H accepts (H^, H^) and H^ halts on input H^
>
> should be the case. So, come clean. Which was the case back then:
>
> (1) H rejects (H^, H^) and H^ does not halt in input H^, or
> (2) H accepts (H^, H^) and H^ halts on input H^, or
> (3) H rejects (H^, H^) and H^ halts on input H^.
>
> Without the comfort blanket of your huge pile of junk x86 code, I
> suspect you won't dare say.
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 is correct ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 is
correct ]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc>
<uMGJI.28030$qk6.2244@fx36.iad> <sd7een$js8$1@gioia.aioe.org>
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From: news.dea...@darjeeling.plus.com (Mike Terry)
Date: Tue, 27 Jul 2021 02:17:31 +0100
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 by: Mike Terry - Tue, 27 Jul 2021 01:17 UTC

On 26/07/2021 16:57, André G. Isaak wrote:
> On 2021-07-26 08:32, olcott wrote:
>
>> You have the paradox incorrectly. While the input to H(P,P) is
>> simulated in pure simulation mode it cannot possibly ever reach a
>> final state thus conclusively proving that this input never halts.
>
> But The halting problem isn't concerned with the input to H(P, P), which
> is simply a representation. It's concerned with the *actual* computation
> which that input represents.  That would be P(P) run independently, and,
> like any other computation, there is exactly one correct answer to the
> question 'does it halt'?
>
> The input to H isn't a computation; it's just data on the tape. So it is
> meaningless to ask whether the input to H halts. What happens inside
> some simulation in some 'mode' isn't the question which the halting
> problem asks.
>
>> Anyone bothering to carefully examine these things must necessarily
>> conclude that the pure simulation of the input to H(P,P) cannot
>> possibly reach its final state. Anyone not bothering to carefully
>> examine these things is a liar and a cheat.
>
> Since a pure simulation of a computation must do *exactly* what the
> actual computation does, if P(P) halts and your simulation does not,
> then it is *not* a pure simulation.
>
>> When H recognizes this infinite behavior pattern and stops simulating
>> its input, this input still never reaches its final state, thus never
>> halts.
>>
>> Many thanks to André G. Isaak for pointing out the proper and best
>> definition of halting we no longer have the ambiguity between halting
>> (reaching its final state) and stopping running (simulation aborted).
>
> The fact that you keep acknowledging me for this is a good illustration
> of the fundamental problem here. I never said anything insightful or
> novel. The definition I gave was exactly the same one you should have
> gotten from reading Linz or Sipser or any other introductory text on the
> subject.
>
> The fact that you are still requiring clarification on something as
> basic as the definition of 'halting' after however many years of working
> on your 'proof' clearly demonstrates that you simply lack the background
> to make any claims at all about this problem.
>
> Now perhaps you should refocus your attention from your 'proof' to
> actually learning the subject which you claim to be talking about. Learn
> what it means to halt. Learn what a computation is (hint: your P is
> definitely *not* a computation). Learn what a proof is (hint: a trace
> does not even remotely qualify as a proof of anything). Then come back
> and review all the work you have done once you've actually learned the
> subject matter.

Sensible advice, but... if PO were /capable/ of learning such topics, he
would have done so many years ago. I'm sure he would have made efforts
in the past (i.e. in his younger life) to take this much easier route,
but I imagine each time he just became baffled by all those symbols, and
abstract ideas like TM, halting, truth, proof, set, function, and so on.
He doesn't do "abstract" because he /can't/ - his brain just doesn't
work that way.

That's why he must denigrate others with the ability to understand basic
concepts like this, as "learned by rote" people. If he acknowledged
that every-day students had basic abilities and understandings that he
simply lacked, he'd be perilously close to having to admit to himself
that he's not the Unacknowledged Genius of his delusions - his whole
world view would come under threat. So he comes up with a way to
dismiss such abilities as "not important" (or better, as actively bad in
some way).

Still, his delusions are harmless enough, and he clearly gets by day to
day, so nothing really /needs/ to be done here... (and I get for some
people arguing with him is fun!) BTW, I think your explanations for PO
are about as clear as it will ever be possible to achieve!

Mike.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

<sdnn91$k7p$1@dont-email.me>

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From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Date: Mon, 26 Jul 2021 19:24:47 -0600
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 by: André G. Isaak - Tue, 27 Jul 2021 01:24 UTC

On 2021-07-26 19:03, olcott wrote:
> On 7/26/2021 7:21 PM, André G. Isaak wrote:
>> On 2021-07-26 18:14, olcott wrote:
>>> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>>>> On 2021-07-26 12:57, olcott wrote:
>>>>
>>>>> In the exact same way it is a mistake to take the fact that the P
>>>>> of int main() { P(P); } definitely reaches its halting state as an
>>>>> indication that it is a halting computation.
>>>>
>>>> You seem to have a serious difficulty processing definitions.
>>>>
>>>> The *definition* of a halting computation is a computation which
>>>> reaches one of its final states in a finite number of steps. This
>>>> refers only to the *actual* computation.
>>>>
>>>> This definition exhaustively partitions the set of computations into
>>>> two mutually-exclusive groups. If we observe that a computation such
>>>> as P(P) reaches a final state in a finite number of steps, it goes
>>>> in the halting group. Otherwise, it goes into the non-halting group.
>>>>
>>>> The definition is clear and unambiguous.
>>>>
>>>> It does not mention anything about simulators or aborted simulations.
>>>>
>>>> It makes no exceptions for 'pathological' inputs.
>>>>
>>>
>>> None-the-less if pathological inputs can be recognized this refutes
>>> Rice.
>>> None-the-less if pathological inputs can be recognized this refutes
>>> Rice.
>>> None-the-less if pathological inputs can be recognized this refutes
>>> Rice.
>>> None-the-less if pathological inputs can be recognized this refutes
>>> Rice.
>>
>> There *are no* patholological inputs. There are simply inputs that H
>> gets wrong. There's nothing absolutely 'pathological' about that.
>
> If my system recognizes whatever you want to call it the cases where the
> halt decider gets wrong, this refutes Rice.

What does that unsubstantiated claim have to do with the question I'd
asked, which was:

How can the claim that H(P, P)==0 be justified *in terms of the
definition of halting* and in terms of the *actual behaviour* of P(P).

P(P) meets this definition. Therefore P(P) halts.

If your your system is really capable of recogonizing the inputs that it
gets wrong, why does it still get them wrong?

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Newsgroups: comp.theory,comp.ai.philosophy,comp.software-eng,sci.math.symbolic
References: <20210719214640.00000dfc@reddwarf.jmc>
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Tue, 27 Jul 2021 01:30 UTC

On 7/26/2021 8:24 PM, André G. Isaak wrote:
> On 2021-07-26 19:03, olcott wrote:
>> On 7/26/2021 7:21 PM, André G. Isaak wrote:
>>> On 2021-07-26 18:14, olcott wrote:
>>>> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>>>>> On 2021-07-26 12:57, olcott wrote:
>>>>>
>>>>>> In the exact same way it is a mistake to take the fact that the P
>>>>>> of int main() { P(P); } definitely reaches its halting state as an
>>>>>> indication that it is a halting computation.
>>>>>
>>>>> You seem to have a serious difficulty processing definitions.
>>>>>
>>>>> The *definition* of a halting computation is a computation which
>>>>> reaches one of its final states in a finite number of steps. This
>>>>> refers only to the *actual* computation.
>>>>>
>>>>> This definition exhaustively partitions the set of computations
>>>>> into two mutually-exclusive groups. If we observe that a
>>>>> computation such as P(P) reaches a final state in a finite number
>>>>> of steps, it goes in the halting group. Otherwise, it goes into the
>>>>> non-halting group.
>>>>>
>>>>> The definition is clear and unambiguous.
>>>>>
>>>>> It does not mention anything about simulators or aborted simulations.
>>>>>
>>>>> It makes no exceptions for 'pathological' inputs.
>>>>>
>>>>
>>>> None-the-less if pathological inputs can be recognized this refutes
>>>> Rice.
>>>> None-the-less if pathological inputs can be recognized this refutes
>>>> Rice.
>>>> None-the-less if pathological inputs can be recognized this refutes
>>>> Rice.
>>>> None-the-less if pathological inputs can be recognized this refutes
>>>> Rice.
>>>
>>> There *are no* patholological inputs. There are simply inputs that H
>>> gets wrong. There's nothing absolutely 'pathological' about that.
>>
>> If my system recognizes whatever you want to call it the cases where
>> the halt decider gets wrong, this refutes Rice.
>
> What does that unsubstantiated claim have to do with the question I'd
> asked, which was:
>
> How can the claim that H(P, P)==0 be justified *in terms of the
> definition of halting* and in terms of the *actual behaviour* of P(P).
>

That does not currently matter, as long as we can consistently use
H(P,P) != P(P) to detect and reject the halting problem counter-examples
we have refuted Rice's theorem.

Is it possible for you to detect when the subject has been changed or do
you have a hard-coded robot brain?

> P(P) meets this definition. Therefore P(P) halts.
>
> If your your system is really capable of recogonizing the inputs that it
> gets wrong, why does it still get them wrong?
>
> André
>

Self-contradictory input is incorrect in the same way that the liar
paradox is neither true nor false.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 is correct ]

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Subject: Re: Black box halt decider is NOT a partial decider [ H(P,P)==0 is correct ]
Newsgroups: comp.theory
References: <20210719214640.00000dfc@reddwarf.jmc> <uMGJI.28030$qk6.2244@fx36.iad> <sd7een$js8$1@gioia.aioe.org> <zyHJI.20655$7H7.13829@fx42.iad> <sd8bim$1set$1@gioia.aioe.org> <87eebrlv2m.fsf@bsb.me.uk> <sdckqo$cm8$1@gioia.aioe.org> <87a6mehx5q.fsf@bsb.me.uk> <sdfbv2$14bi$2@gioia.aioe.org> <875yx0he2s.fsf@bsb.me.uk> <sdffqm$jsh$1@gioia.aioe.org> <87zgucfux4.fsf@bsb.me.uk> <sdi9vb$r9b$1@dont-email.me> <87eebnfc8c.fsf@bsb.me.uk> <19Kdna-u6-AOSWH9nZ2dnUU78QXNnZ2d@brightview.co.uk> <87tukjdqmi.fsf@bsb.me.uk> <sdjlo4$1oct$1@gioia.aioe.org> <eOadnfv1CNxEEGD9nZ2dnUU78RPNnZ2d@brightview.co.uk> <4826ab33-061b-472e-a1a5-e2ded35ecd82n@googlegroups.com> <Ob2dneXfOsPHVGD9nZ2dnUU78aHNnZ2d@brightview.co.uk> <tOudnQr4N_JfUGD9nZ2dnUU78fHNnZ2d@brightview.co.uk> <877dhec8wh.fsf@bsb.me.uk> <BM6dnZyWXYxlYWD9nZ2dnUU78WfNnZ2d@brightview.co.uk> <dtudnUPpgO0PWmP9nZ2dnUU7-TnNnZ2d@giganews.com> <sdmm0d$1n1$1@dont-email.me> <xbednQcGH_k2w2L9nZ2dnUU78IHNnZ2d@brightview.co.uk>
From: NoO...@NoWhere.com (olcott)
Date: Mon, 26 Jul 2021 20:41:13 -0500
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 by: olcott - Tue, 27 Jul 2021 01:41 UTC

On 7/26/2021 8:17 PM, Mike Terry wrote:
> On 26/07/2021 16:57, André G. Isaak wrote:
>> On 2021-07-26 08:32, olcott wrote:
>>
>>> You have the paradox incorrectly. While the input to H(P,P) is
>>> simulated in pure simulation mode it cannot possibly ever reach a
>>> final state thus conclusively proving that this input never halts.
>>
>> But The halting problem isn't concerned with the input to H(P, P),
>> which is simply a representation. It's concerned with the *actual*
>> computation which that input represents.  That would be P(P) run
>> independently, and, like any other computation, there is exactly one
>> correct answer to the question 'does it halt'?
>>
>> The input to H isn't a computation; it's just data on the tape. So it
>> is meaningless to ask whether the input to H halts. What happens
>> inside some simulation in some 'mode' isn't the question which the
>> halting problem asks.
>>
>>> Anyone bothering to carefully examine these things must necessarily
>>> conclude that the pure simulation of the input to H(P,P) cannot
>>> possibly reach its final state. Anyone not bothering to carefully
>>> examine these things is a liar and a cheat.
>>
>> Since a pure simulation of a computation must do *exactly* what the
>> actual computation does, if P(P) halts and your simulation does not,
>> then it is *not* a pure simulation.
>>
>>> When H recognizes this infinite behavior pattern and stops simulating
>>> its input, this input still never reaches its final state, thus never
>>> halts.
>>>
>>> Many thanks to André G. Isaak for pointing out the proper and best
>>> definition of halting we no longer have the ambiguity between halting
>>> (reaching its final state) and stopping running (simulation aborted).
>>
>> The fact that you keep acknowledging me for this is a good
>> illustration of the fundamental problem here. I never said anything
>> insightful or novel. The definition I gave was exactly the same one
>> you should have gotten from reading Linz or Sipser or any other
>> introductory text on the subject.
>>
>> The fact that you are still requiring clarification on something as
>> basic as the definition of 'halting' after however many years of
>> working on your 'proof' clearly demonstrates that you simply lack the
>> background to make any claims at all about this problem.
>>
>> Now perhaps you should refocus your attention from your 'proof' to
>> actually learning the subject which you claim to be talking about.
>> Learn what it means to halt. Learn what a computation is (hint: your P
>> is definitely *not* a computation). Learn what a proof is (hint: a
>> trace does not even remotely qualify as a proof of anything). Then
>> come back and review all the work you have done once you've actually
>> learned the subject matter.
>
> Sensible advice, but... if PO were /capable/ of learning such topics, he
> would have done so many years ago.  I'm sure he would have made efforts
> in the past (i.e. in his younger life) to take this much easier route,
> but I imagine each time he just became baffled by all those symbols, and
> abstract ideas like TM, halting, truth, proof, set, function, and so on.
>  He doesn't do "abstract" because he /can't/ - his brain just doesn't
> work that way.
>
> That's why he must denigrate others with the ability to understand basic
> concepts like this, as "learned by rote" people.

It is clearly true that mathematicians never ever carefully examine the
philosophical underpinnings of the fundamental nature of truth itself as
proven by the fact the no one will evaluate the Tarski proof.
http://www.liarparadox.org/Tarski_Proof_275_276.pdf

They never realize that these philosophical underpinnings can turn the
ultimate basis of Gödel's work on its ear because they never examine
these things.

If analytical truth and provability mutually define each other then true
and unprovable is utterly impossible and incompleteness cannot possibly
exist. Sentences that are neither provable nor refutable are simply not
truth bearers.

> If he acknowledged
> that every-day students had basic abilities and understandings that he
> simply lacked, he'd be perilously close to having to admit to himself
> that he's not the Unacknowledged Genius of his delusions - his whole
> world view would come under threat.  So he comes up with a way to
> dismiss such abilities as "not important" (or better, as actively bad in
> some way).
>
> Still, his delusions are harmless enough, and he clearly gets by day to
> day, so nothing really /needs/ to be done here...  (and I get for some
> people arguing with him is fun!)  BTW, I think your explanations for PO
> are about as clear as it will ever be possible to achieve!
>
>
> Mike.
>

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

<sdnrb7$al5$1@dont-email.me>

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https://www.novabbs.com/devel/article-flat.php?id=18994&group=comp.theory#18994

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Path: i2pn2.org!i2pn.org!eternal-september.org!reader02.eternal-september.org!.POSTED!not-for-mail
From: agis...@gm.invalid (André G. Isaak)
Newsgroups: comp.theory
Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Date: Mon, 26 Jul 2021 20:34:14 -0600
Organization: Christians and Atheists United Against Creeping Agnosticism
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 by: André G. Isaak - Tue, 27 Jul 2021 02:34 UTC

On 2021-07-26 19:30, olcott wrote:
> On 7/26/2021 8:24 PM, André G. Isaak wrote:
>> On 2021-07-26 19:03, olcott wrote:
>>> On 7/26/2021 7:21 PM, André G. Isaak wrote:
>>>> On 2021-07-26 18:14, olcott wrote:
>>>>> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>>>>>> On 2021-07-26 12:57, olcott wrote:
>>>>>>
>>>>>>> In the exact same way it is a mistake to take the fact that the P
>>>>>>> of int main() { P(P); } definitely reaches its halting state as
>>>>>>> an indication that it is a halting computation.
>>>>>>
>>>>>> You seem to have a serious difficulty processing definitions.
>>>>>>
>>>>>> The *definition* of a halting computation is a computation which
>>>>>> reaches one of its final states in a finite number of steps. This
>>>>>> refers only to the *actual* computation.
>>>>>>
>>>>>> This definition exhaustively partitions the set of computations
>>>>>> into two mutually-exclusive groups. If we observe that a
>>>>>> computation such as P(P) reaches a final state in a finite number
>>>>>> of steps, it goes in the halting group. Otherwise, it goes into
>>>>>> the non-halting group.
>>>>>>
>>>>>> The definition is clear and unambiguous.
>>>>>>
>>>>>> It does not mention anything about simulators or aborted simulations.
>>>>>>
>>>>>> It makes no exceptions for 'pathological' inputs.
>>>>>>
>>>>>
>>>>> None-the-less if pathological inputs can be recognized this refutes
>>>>> Rice.
>>>>> None-the-less if pathological inputs can be recognized this refutes
>>>>> Rice.
>>>>> None-the-less if pathological inputs can be recognized this refutes
>>>>> Rice.
>>>>> None-the-less if pathological inputs can be recognized this refutes
>>>>> Rice.
>>>>
>>>> There *are no* patholological inputs. There are simply inputs that H
>>>> gets wrong. There's nothing absolutely 'pathological' about that.
>>>
>>> If my system recognizes whatever you want to call it the cases where
>>> the halt decider gets wrong, this refutes Rice.
>>
>> What does that unsubstantiated claim have to do with the question I'd
>> asked, which was:
>>
>> How can the claim that H(P, P)==0 be justified *in terms of the
>> definition of halting* and in terms of the *actual behaviour* of P(P).
>>
>
> That does not currently matter,

Of course it currently matters.

> as long as we can consistently use
> H(P,P) != P(P) to detect and reject the halting problem counter-examples
> we have refuted Rice's theorem.

How on earth would that refute Rice's theorem?

I *think* that by H(P,P) != P(P) you're *trying* (badly) to say 'the
answer given by H(P, P) does not match the halting behaviour of P(P).

But *we* would be the ones recognizing these counterexamples. That your
machine gets these cases wrong means the machine certainly isn't
recognizing them.

> Is it possible for you to detect when the subject has been changed or do
> you have a hard-coded robot brain?

You seem to 'change the subject' whenever you get stuck. Why not try
actually sticking with a subject for a change rather than simply
sweeping all objections under the rug and then claiming that no one has
given you good arguments (a common tact with you).

>> If your your system is really capable of recogonizing the inputs that
>> it gets wrong, why does it still get them wrong?

> Self-contradictory input is incorrect in the same way that the liar
> paradox is neither true nor false.

There *is no input* when I ask you to justify the claim that P(P)
doesn't halt when it clearly does not. I'm asking about the behaviour of
the computation P(P).

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

Re: Black box halt decider is NOT a partial decider [ H refutes Rice's Theorem ]

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https://www.novabbs.com/devel/article-flat.php?id=18995&group=comp.theory#18995

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Subject: Re: Black box halt decider is NOT a partial decider [ H refutes
Rice's Theorem ]
Newsgroups: comp.theory
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Tue, 27 Jul 2021 02:36 UTC

On 7/26/21 6:30 PM, olcott wrote:
> On 7/26/2021 8:24 PM, André G. Isaak wrote:
>> On 2021-07-26 19:03, olcott wrote:
>>> On 7/26/2021 7:21 PM, André G. Isaak wrote:
>>>> On 2021-07-26 18:14, olcott wrote:
>>>>> On 7/26/2021 2:55 PM, André G. Isaak wrote:
>>>>>> On 2021-07-26 12:57, olcott wrote:
>>>>>>
>>>>>>> In the exact same way it is a mistake to take the fact that the P
>>>>>>> of int main() { P(P); } definitely reaches its halting state as
>>>>>>> an indication that it is a halting computation.
>>>>>>
>>>>>> You seem to have a serious difficulty processing definitions.
>>>>>>
>>>>>> The *definition* of a halting computation is a computation which
>>>>>> reaches one of its final states in a finite number of steps. This
>>>>>> refers only to the *actual* computation.
>>>>>>
>>>>>> This definition exhaustively partitions the set of computations
>>>>>> into two mutually-exclusive groups. If we observe that a
>>>>>> computation such as P(P) reaches a final state in a finite number
>>>>>> of steps, it goes in the halting group. Otherwise, it goes into
>>>>>> the non-halting group.
>>>>>>
>>>>>> The definition is clear and unambiguous.
>>>>>>
>>>>>> It does not mention anything about simulators or aborted simulations.
>>>>>>
>>>>>> It makes no exceptions for 'pathological' inputs.
>>>>>>
>>>>>
>>>>> None-the-less if pathological inputs can be recognized this refutes
>>>>> Rice.
>>>>> None-the-less if pathological inputs can be recognized this refutes
>>>>> Rice.
>>>>> None-the-less if pathological inputs can be recognized this refutes
>>>>> Rice.
>>>>> None-the-less if pathological inputs can be recognized this refutes
>>>>> Rice.

Nope, because just because H recognises that it will be wrong, doesn't
let it give the real right answer, all it can do is admit that it will
be wrong. Rice is still upheld.

>>>>
>>>> There *are no* patholological inputs. There are simply inputs that H
>>>> gets wrong. There's nothing absolutely 'pathological' about that.
>>>
>>> If my system recognizes whatever you want to call it the cases where
>>> the halt decider gets wrong, this refutes Rice.
>>
>> What does that unsubstantiated claim have to do with the question I'd
>> asked, which was:
>>
>> How can the claim that H(P, P)==0 be justified *in terms of the
>> definition of halting* and in terms of the *actual behaviour* of P(P).
>>
>
> That does not currently matter, as long as we can consistently use
> H(P,P) != P(P) to detect and reject the halting problem counter-examples
> we have refuted Rice's theorem.

Except that if P converts this new 'invalid' result to Halting (or
Non-Halting) behavior, then H(P,P) is still != P(P), as Invalid is not
the same as Halting (or non-halting). H is STILL wrong, even if it
announces that it knows it will be wrong.

And, as long as H does return an answer (or can be proved to never
return an answer) then another decider H1, can get the right answer for
P(P) by doing H1(P,P), so it isn't like P doesn't have a right answer.

>
> Is it possible for you to detect when the subject has been changed or do
> you have a hard-coded robot brain?
>

Also, you STILL haven't retuted Rice, as you still haven't shown that H
got the right answer.

>> P(P) meets this definition. Therefore P(P) halts.
>>
>> If your your system is really capable of recogonizing the inputs that
>> it gets wrong, why does it still get them wrong?
>>
>> André
>>
>
> Self-contradictory input is incorrect in the same way that the liar
> paradox is neither true nor false.
>

WRONG. For any specific H that answers H(P,P) then there IS a right
answer for "Does P(P) Halt?". Thus the proper question for the Halting
Problem is NOT a contradiction.

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